Use de-Moivre's theorem to find the reciprocal of each number below. $$\sqrt 3 - i$$

Given $\sqrt{3}-i$ , we need to find the reciprocal of it using de-Moivre's theorem.

$$\frac{1}{\sqrt 3-i} $$

$$= \frac{1(\cos0^c + i\sin 0^c)}{2\big(\frac{\sqrt3}{2}-\frac{i}{2}\big)}$$

$$= \frac{1}{2}\cdot \frac{\cos\big(\frac{\pi}{6}\big) - i\sin\big(\frac{\pi}{6})}{\big(\frac{\sqrt3}{2}-\frac{i}{2}\big)}$$

$$ = \frac{1}{2}\cdot \cos\biggr(\frac{\pi}{6}\biggr) - i\sin\biggr(\frac{\pi}{6}\biggr)$$

$$ = \frac{1}{2} \cdot \biggr(\frac{\sqrt3}{2} +i \frac{1}{2}\biggr )$$

$$\boxed { = \frac{\sqrt3}{4}+\frac{i}{4}}$$

Does my assumption seem correct?

  • 1
    $\begingroup$ I see a degrees symbol. Note that in complex radians are always used $\endgroup$ – Rhys Hughes May 11 '18 at 10:30
  • $\begingroup$ @RhysHughes Pardon me, what did you mean by that? $\endgroup$ – Busi May 11 '18 at 10:34
  • $\begingroup$ When working with complex numbers, and dealing with angles, it is standard to use radians to measure them instead of degrees. I noticed $\cos(0^0)$ in your question, when it should be $\cos (0^c)$ or $0 rad$ $\endgroup$ – Rhys Hughes May 11 '18 at 10:52

De Moivre's Theorem states that: $$[r(\cos \theta + i \sin \theta)]^n=r^n(\cos (n\theta)+ i \sin (n\theta))$$

To find the reciprocal, take $n=-1$. $$z=\sqrt{3}-i\to r=2, \theta=\tan^{-1}\bigg({\frac{-1}{\sqrt{3}}}\bigg)=\frac{-\pi}{6}$$ Hence $z^{-1}=2^{-1}(\cos{(-1)(\frac{-\pi}{6})}+ i\sin{(-1)(\frac{-\pi}{6})})=\frac{1}{2}(\cos \frac{\pi}{6}+i\sin\frac{\pi}{6})$

$$=\frac{\sqrt{3}}{4}+\frac{1}{4}i$$ as you achieved.


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